Solution:

Since the area of square $ABCD$ is $196 = 14^2~(\mathrm{cm}^2)$, then the side of square $ABCD$ is $14~\mathrm{cm}$.

We claim that $\triangle ABE$ is equilateral. To prove this, we make a reverse construction, starting from an equilateral $\triangle ABE'$, building up to square $ABCD$, and eventually showing that points $E$ and $E'$ coincide. Thus,

Step 1. Let $E'$ be a point inside square $ABCD$ such that $\triangle ABE'$ is equilateral.
Hence, in particular, $\angle ABE' = 60^\circ$.

Step 2. Then $\angle E'BC = 90^\circ - 60^\circ = 30^\circ$.

Step 3. Since $BE' = AB = BC$ ($\triangle ABE'$ is equilateral and $ABCD$ is a square), we conclude that $\triangle CE'B$ is isosceles with $BE' = BC$ and $\angle E'BC = 30^\circ$.

Step 4. This in turn implies that both base angles of $\triangle CE'B$ are $\frac{1}{2}(180^\circ - 30^\circ) = 75^\circ$. In particular, $\angle BCE' = 75^\circ$. (Note: Since $\angle BCE' < 90^\circ$, point $E'$ is between parallel lines $AB$ and $CD$, and hence $E'$ is indeed inside square $ABCD$, and not "above" $CD$ or outside $ABCD$.)

Step 5. But then $\angle DCE' = 90^\circ - 75^\circ = 15^\circ$.

Step 6. Analogously (or by symmetry) we can show that $\angle CDE' = 15^\circ$.

Step 7. This means that $\triangle DCE'$ is isosceles with $DE' = CE'$ and base $\angle DE'C = 150^\circ$.

Step 8. Thus, $E'$ is at the same distances from $C$ and $D$, $\angle DE'C = 150^\circ$, and $E'$ is inside square $ABCD$. But there is only one such point with all these properties, namely, the given point $E$. We conclude that point $E'$ is, after all, the same as point $E$.

To finish off the problem, we use the fact that $\triangle ABE'$ is equilateral by construction; i.e., the original $\triangle ABE$ is equilateral. Hence its perimeter is equal to $3 \cdot AB = 3 \cdot 14 = 42~\mathrm{cm}$.

Solution:

Our goal will be to confirm that $\angle EBC = 30^\circ$. For those who know some trigonometry, recall a formula for $\tan$ (half-angle):
$$\tan^2 \alpha = \frac{\cos^2 \alpha}{\sin^2 \alpha} = \frac{\frac{1}{2}(1-\cos 2\alpha)}{\frac{1}{2}(1+\cos 2\alpha)} \Rightarrow \tan^2 15^\circ = \frac{1-\cos 30^\circ}{1+\cos 30^\circ} = \frac{1-\frac{\sqrt{3}}{2}}{1+\frac{\sqrt{3}}{2}} = \frac{2-\sqrt{3}}{2+\sqrt{3}}.$$
With the help of a little bit of algebra, we rationalize the denominator and obtain:
$$\frac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})} = \frac{(2-\sqrt{3})^2}{2^2-3} = \frac{(2-\sqrt{3})^2}{1} \Rightarrow \tan 15^\circ = +\sqrt{(2-\sqrt{3})^2} = 2-\sqrt{3} (>0).$$
Now back to our problem. Drop a perpendicular from point $E$ to side $BC$ of the square and mark by $H$ the foot of this perpendicular. Since $E$ is at the same distances from $C$ and $D$, it easily follows that $E$ is half-way between parallel lines $AD$ and $BC$; i.e., $EH = \frac{1}{2} AB = 7~\mathrm{cm}$.

Just as above, from isosceles $\triangle DEC$ we know that $\angle ECD = 15^\circ$. This means that $\angle CEH = 15^\circ$ ($EH \parallel DC$, $CE$ is a transversal, and $\angle ECD$ and $\angle CEH$ are alternating interior angles). Thus, from right $\triangle EHC$ we can calculate $CH = 7 \tan 15^\circ$. This implies that $BH = BC - CH = 14 - 7 \tan 15^\circ$.

We are ready to calculate $\cot \angle EBC$ from right $\triangle EHB$:
$$\cot \angle EBC = \cot \angle EBH = \frac{BH}{EH} = \frac{14 - 7 \tan 15^\circ}{7} = 2 - \tan 15^\circ = 2 - (2-\sqrt{3}) = \sqrt{3}.$$
But $\cot 30^\circ = \sqrt{3}$, so the acute $\angle EBC$ must be $30^\circ$. We proved what we aimed for!

To finish off the problem, note that $\angle ABE = 90^\circ - \angle EBC = 90^\circ - 30^\circ = 60^\circ$. By symmetry, $\angle ABE = 60^\circ$ and $\triangle ABE$ is equilateral. Thus, its perimeter is $3 AB = 42~\mathrm{cm}$.

Solution:

It is not hard to compute the ratios of a 15-75-90 right triangle without any trig, provided one knows the ratios of the 30-60-90 triangle. Just divide the 75-degree angle into 15-degree and 60-degree angles, dissecting the original triangle into a 15-15-150 and 30-60-90. Quite elementary from there.